Solution
= Solution
Yes. The old <matrix> is triangular, with <determinant> zero and <trace> $0+0+1+1=2$. The upper-left two-by-two block of the new <matrix> has <determinant> zero, so the whole <determinant> remains zero. Its <trace> is
$$
-\frac12+\frac12+\frac32+\frac12=2.
$$
Thus
$$
\boxed{\det[F]_{\mathcal B}=\det[F]_{\mathcal C}=0,\qquad
\operatorname{tr}[F]_{\mathcal B}=\operatorname{tr}[F]_{\mathcal C}=2.}
$$
This verifies both <change of basis> invariants from part (b).